# Is this the stochastic differential of the given process?

Let the process $$X_t:=\cos(B_{1,t}B_{2,t}),$$ where $$(B_1, B_2)$$ is a bi-dimensional correlated Brownian motion, be given. Is the following the correct Ito representation of its stochastic differential? $$dX_t = \\-\frac12 \left\{\rho^{1,1}B_{2,t}^2\cos{(B_{1,t}B_{2,t})} + \rho^{2,2}B_{1,t}^2\cos{(B_{1,t}B_{2,t})} \\+ \rho^{2,1}[\sin{(B_{1,t}B_{2,t})} + B_{1,t}B_{2,t}\cos{(B_{1,t}B_{2,t})}]\right\}dt \\- B_{2,t}\sin{(B_{1,t}B_{2,t})}dB_{1,t}-B_{1,t}\sin{(B_{1,t}B_{2,t})}dB_{2,t}$$

Let $$Y_t:=B_{1,t}B_{2,t}$$, so that $$X_t=\cos(Y_t)$$ and $$dX_t=-\sin(Y_t)dY_t -{1\over 2}\cos(Y_t)d\langle Y\rangle_t.$$ To make this more explicit, use $$dY_t=B_{1,t} dB_{2,t}+B_{2,t} dB_{1,t}+\rho^{2,1} dt$$ (if I guess correctly at what you mean by $$\rho^{2,1}$$), and $$d\langle Y\rangle_t = B_{1,t}^2\rho^{2,2} dt +B_{2,t}^2\rho^{1,1} dt + 2B_{1,t}B_{2,t}\rho^{2,1} dt.$$ It looks to me like your formula is correct except your $$-{1\over 2}\rho^{2,1}\sin(B_{1,t}B_{2,t}) dt$$ term shouldn't have a $${1\over 2}$$.